This question got mixed up in translation from another document. For illustration, I have negated each one. These are not the original questions that you were asked.
∃x( x ≤ 1 )
∀x( x ≥ 0 )
∀x( x ≤ −2 ∨ 3 ≤ x ) (Be careful. Notation (−2 < x < 3) is shorthand for (−2 < x ∧ x < 3). You need to apply DeMorgan's law, converting the ∧ to ∨.
∃x∃yQ(x, y)
∃x(Q(x, "Jeopardy") ∧ Q(x, "Wheel of Fortune"))
∀y∃xQ(x, y)
∃x∃y(x ≠ y ∧ Q(x, "Wheel of Fortune") ∧ Q(y, "Wheel of Fortune"))